Answer
$$\ln \left|\frac{\sqrt{x^{2}+16}}{4}+\frac{x}{4}\right|+C$$
Work Step by Step
Given $$\int \frac{d x}{\sqrt{16+x^{2}}}$$
Let
$$x=4\tan u \ \ \ \ dx=4\sec^2 udu $$
Then
\begin{align*}
\int \frac{d x}{\sqrt{16+x^{2}}}&=\int \frac{4\sec^2 udu}{\sqrt{16+16\tan^{2}u}}\\
&=\int \frac{4\sec^2 udu}{4\sec u}\\
&=\int \sec udu\\
&=\ln |\sec u+\tan u|+C\\
&= \ln \left|\frac{\sqrt{x^{2}+16}}{4}+\frac{x}{4}\right|+C
\end{align*}
